Al1+++3e−⟶Al $\begin{aligned}
\text { Moles of electron } & =\frac{2 \times 30 \times 60}{96500} \
& =\frac{36}{965}
\end{aligned}\begin{aligned}
\text { Moles of } \mathrm{Al} & =\frac{36}{3 \times 965} \
& =\frac{12}{965}
\end{aligned}\text { Mass of } \mathrm{Al}=\frac{12}{965} \times 27=0.336 \mathrm{gm}$